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Posted by on Sunday, February 27, 2011 at 3:42pm.

The Ka value for ethanoic acid is 1.8x10^-5, what would be the pH of a 2.0x10^-2 M solution of ethanoic acid?

  • Chemistry - , Sunday, February 27, 2011 at 7:01pm

    ............CH3COOH ==> H^+ + CH3COO^-
    initial.....0.02M........0.......0
    change........-x.........+x......+x
    final......0.02-x.........x.......x

    Ka = (H^+)(CH3COO^-)/((CH3COOH)
    Plug the ICE chart values into the Ka expression, solve for H, then convert to pH.

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